A solid ball of radius \(R\) has a charge density \(\rho \) given by \(\rho = {\rho _0}\left( {1 - \frac{r}{R}} \right)\) for \(0 \leq r \leq R\). The electric field outside the ball is

A \(\frac{{{\rho _0}{R^3}}}{{{\varepsilon _0}{r^2}}}\)
B \(\frac{{{4\rho _0}{R^3}}}{{{3\varepsilon _0}{r^2}}}\)
C \(\frac{{{3\rho _0}{R^3}}}{{{4\varepsilon _0}{r^2}}}\)
D \(\frac{{{\rho _0}{R^3}}}{{{12\varepsilon _0}{r^2}}}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{{{\rho _0}{R^3}}}{{{12\varepsilon _0}{r^2}}}\)

Step-by-step Solution

Detailed explanation

\(\text { Charge density, } \rho=\rho_{0}\left(1-\frac{r}{R}\right)\) \(d q=\rho d v\) \(q_{i n}=\int d q=\rho d v\) \(=\rho_{0}\left(1-\frac{r}{R}\right) 4 \pi r^{2} d r \quad\left(\because d v=4 \pi \mathrm{r}^{2} \mathrm{dr}\right)\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From JEE Mains

Explore more questions on app